chase_manhattan_bank

Case Study: Chase Manhattan Bank

Name of the group members ID Number

Alexandra CERISIER 5472100519Chusak Deetrakunwattanapoen 7472100374

Gopi Krishna Pamula 7462100509Mohammed Faruk Ahmed 7472100333

Nam Doc Tran 7472100382Sukana Sasiwachiranggool 7472100291

Varavut Nitistaporn 7472100127Headings: Scenario in brief

Problem identification

Decision Variables, Objective function, Constraints

LP model

Plot the model in spread sheet to obtain optimum solution

Focus on discussion questions

Scenario in brief: Chase Manhattan Bank in New York is facing employee-scheduling problem due to insufficient manpower in pick hours of the operation. To meet the requirements of having sufficient personnel in every hour in most cost effective way, it is required to have some part time personnel in different hours as per demand. There is another option to have overtime. There are some data with constraints given in the case study based on which we have to find the best labor schedule with the minimum cost.

Identified Problem:

In some busy hours,

Number of fulltime personnel < Required number of personnel

Decision Variables and Objective Function:Our Assumptions:

1. We assume that the total number of Full-time personnel = F. We also assume that full-time personnel work from 9 am to 5 pm. If they work between 5 pm and 7 pm, it will be counted as over-time.

2. Let’s consider the part-time and overtime personnel as below table 1 to meet the required workforce in different specified hours. To satisfy the constrain No. 3, every part-timer will work at least 4 hours. We consider every part-timer will work exactly 4 hours to simplify calculation, which obviously satisfy the Constraint No. 3.

Lets assume that

F= Full time personnelP1 is the part-timers join at 9 A.M. P2 is the part-timers join at 10 A.M. P3 is the part-timers join at 11 A.M. P4 is the part-timers join at 12 Noon P5 is the part-timers join at 1 P.M. P6 is the part-timers join at 2 P.M.

OT1 is the full-timers work during 5 – 6 P.M as overtime. OT2 is the full-timers work during 6 – 7 P.M as overtime.F, P1, P2, P3, P4, P5, P6, OT1, OT2 >= 0Since required number is increased until 2-3 P.M. period, let’s add and reduce the part-timers (Pi = P1, P2.......P6) as table 1. We also assume that we may need over-timer in last 2 time periods 5-6 and 6-7

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P.M. and they are OT1 and OT2. Actually this mathematical model will give us the closer values of Pi, OT1, OT2 and F from where optimum schedule can be found by solving in spread sheet.

TABLE 1

TIME PERIOD

FULL TIMER

OVER-TIMER PART-TIMER NUMBER OF PERSONNEL REQUIRED

9-10 A.M. F 0 P1 14

10-11 F 0 P1+P2 25

11-12 0.5F 0 P1+P2+P3 26

12-1 P.M. 0.5F 0 P1+P2+P3+P4 38

1-2 F 0 P2+P3+P4+P5 55

2-3 F 0 P3+P4+P5+P6 60

3-4 F 0 P4+P5+P6 51

4-5 F 0 P5+P6 29

5-6 0 OT1 P6 14

6-7 0 OT2 0 9

Total 321Equation from table 1.

TABLE 2

Time period Equations Equation No.

9-10 A.M. F+P1 >= 14 1

10-11 F+P1+P2 >= 25 2

11-12 0.5F+P1+P2+P3 >= 26 3

12-1 P.M. 0.5F+P1+P2+P3+P4 >= 38 4

1-2 F+P2+P3+P4+P5 >= 55 5

2-3 F+P3+P4+P5+P6 >= 60 6

3-4 F+P4+P5+P6 >= 51 7

4-5 F+P5+P6 >= 29 8

5-6 OT1+P6 >= 14 9

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6-7 OT2 >= 9 10

Objective Function: Minimum cost to meet personnel requirements by optimum workforce schedule.

Minimum Cost = 10.11(F+F+.5F+.5F+F+F+F+F)+7.82*4 (P1+P2+P3+P4+P5+P6) +8.08(OT1+OT2)

Or, Minimum Cost = 10.11*7F + 7.82*4 (P1+P2+P3+P4+P5+P6) +8.08(OT1+OT2)Or, Minimum Cost = 70.77F + 31.28(P1+P2+P3+P4+P5+P6) +8.08(OT1+OT2)

Constraints:Constraint 1:To get optimum solution (min cost), we consider the highest value of part-timer hours (40% of the total personnel) as part-timers are paid least. 4(P1+P2+P3+P4+P5+P6) = 40% of (14+25+26+38+55+60+51+29+14+9) OR, 4(P1+P2+P3+P4+P5+P6) = 0.4 * 321 OR, (P1+P2+P3+P4+P5+P6) = (0.4*321)/4 OR, (P1+P2+P3+P4+P5+P6) = 32.1 (~32) -----------------------------------------(11)Constraint 2:Applied in Equation 1 to Equation 8 of Table 2Constraint 3:Applied in Equation 1 to Equation 9 of Table 2, We have assumed that every part-timer will work exactly 4 hours for the simplicity of the model but it does not violate any of the constraints.Constraint 4:Applied in Equation 3 to Equation 4 of Table 2Constraint 5:Applied in Equation 9 to Equation 10 of Table 2Constraint 6:Daily overtime,OT1+OT2 <= 2F

Weekly Overtime,5*(OT1+OT2) <= 5FNon-negativity: F, P1, P2, P3, P4, P5, P6, OT1, OT2 >= 0Calculations:From (3), P1+P2+P3=26 - 0.5F -----------------(12)From (7), P4+P5+P6=51 – F --------------------(13)Apply (12) and (13) in equation (11),26 – 0.5F + 51 – F = 321.5F = 45F = 30 = Full-timer --------------------------------(14)We can find the following values by calculating from equation 1 throughP1 and P2 = 0 (Since the Full-timer is 30 which is greater than the required personnel in the respective hours.)P3 = 11, P4 = 12, P5 = 2 and P6 = 5, OT1 = 9 and OT2 = 9

Plot the model in spreadsheet to obtain solution:We plot the LP model as the following table (Table 3) to solve the problem.

Table 3

Chase Manhattan Bank Required Hours

3

Full Time Personnel OT1 OT2 P1 P2 P3 P4 P5 P6

Part-timers LHS Operator RHS

Cost (Min) 10.11 8.08 8.08 7.82 7.82 7.82 7.82 7.82 7.82 7.82 2264.38

9:00-10:00 1 0 0 0 0 0 0 0 0 0 30 >= 1410:00-11:00 1 0 0 0 0 0 0 0 0 0 30 >= 2511:00-12:00 0.5 0 0 0 0 1 0 0 0 11 16 >= 2612:00-13:00 0.5 0 0 0 0 1 1 0 0 23 17 >= 3813:00-14:00 1 0 0 0 0 1 1 1 0 25 33 >= 5514:00-15:00 1 0 0 0 0 1 1 1 1 30 34 >= 6015:00-16:00 1 0 0 0 0 0 1 1 1 19 33 >= 5116:00-17:00 1 0 0 0 0 0 0 1 1 7 32 >= 2917:00-18:00 0 1 0 0 0 0 0 0 1 5 2 >= 1418:00-19:00 0 0 1 0 0 0 0 0 0 0 1 >= 9Solution Values 30 1 1 1 1 1 1 1 1 120 Total Hours 210 1 1 0 0 4 4 4 4 228 Total Part-time 16

After applying all constraints in spreadsheet, we have got the following solution Please see the answer sheet attached herewith for detail.

Table 4

Chase Manhattan Bank Required Hours


Part-timers LHS Operator RHS

Cost (Min) 10.11 8.08 8.08 7.82 7.82 7.82 7.82 7.82 7.82 7.82 3253.34

9:00-10:00 1 0 0 0 0 0 0 0 0 0 30 >= 1410:00-11:00 1 0 0 0 0 0 0 0 0 0 30 >= 2511:00-12:00 0.5 0 0 0 0 1 0 0 0 11 26 >= 2612:00-13:00 0.5 0 0 0 0 1 1 0 0 23 40 >= 3813:00-14:00 1 0 0 0 0 1 1 1 0 25 55 >= 5514:00-15:00 1 0 0 0 0 1 1 1 1 30 62 >= 6015:00-16:00 1 0 0 0 0 0 1 1 1 19 51 >= 5116:00-17:00 1 0 0 0 0 0 0 1 1 7 37 >= 2917:00-18:00 0 1 0 0 0 0 0 0 1 5 14 >= 1418:00-19:00 0 0 1 0 0 0 0 0 0 0 9 >= 9Solution Values 30 7 9 0 0 11 14 0 7 120 Total Hours 210 7 9 0 0 44 56 0 28 354 Total Part-time 128

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Discussion Questions:

1. Total minimum cost is $3253.34. Minimum cost schedule is given below.


9:00-10:00 30

10:00-11:00 30

11:00-12:00 15 11

12:00-13:00 15 11 14

13:00-14:00 30 11 14

14:00-15:00 30 11 14 7

15:00-16:00 30 14 7

16:00-17:00 30 7

17:00-18:00 7 7

18:00-19:00 9

2. Limitations of the model used to answer question no. 1 are – Part-timers must be less than 40% of the total workforce. Full-timers are allowed to work 8 hours including 1 hour lunch time Part-timers work at least 4 hours and maximum 7 hours without any lunch break. Fifty percent full timers go to lunch in the periods of 11-12 AM and 12-1 PM. Every full-timers may have 2 hours over-time per day and 5 hours per week if necessary.

3. If we would change the value 40% to higher for the part-timer, it will have significant impact on the solution that we have found. Because –

If we can have more than 40% part-timers of total required personnel, number of full-timers would be less than what we found (F=30) which would reduce the cost significantly. Part-timers cost less than full-timers. Other issue is that we have full-timers more than our requirement in some periods. This could be minimized if we could consider higher percentage of part-timers that would result low number of full-timers.

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